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Units, Constants and Useful Formulas
Complex Scalar Fields
---------------------
Think of 2 real scalar fields φ_{1} and φ_{2} and combine them together
in the following manner.
φ = φ_{1} + iφ_{2} and φ^{*} = φ_{1} - iφ_{2}
We can write that ∂φ/∂x^{μ} = ∂φ_{1}/∂x^{μ} + i∂φ_{2}/∂x^{μ}
.
L = φ_{1}^{2}/2 - (∂_{x}φ_{1}/2)^{2} - (∂_{y}φ_{1}/2)^{2} - (∂_{z}φ_{1}/2)^{2} - m^{2}φ_{1}^{2}
.
+ φ_{2}^{2}/2 - (∂_{x}φ_{2}/2)^{2} - (∂_{y}φ_{2}/2)^{2} - (∂_{z}φ_{2}/2)^{2} - m^{2}φ_{2}^{2}
This can be simplified to:
..
= φφ^{*}/2 - ∂_{x}φ^{*}∂_{x}φ - ∂_{y}φ^{*}∂_{y}φ - ∂_{z}φ^{*}∂_{x}φ - m^{2}φ^{*}φ/2
It is conventional to drop the 1/2. Therefore, in abbreviated form,
we can write:
L = ∂_{μ}φ∂^{μ}φ^{*} - m^{2}φ^{*}φ
The quantity m^{2}φ^{*}φ is referred to as the 'mass term'.
Now apply a phase transformation, exp(iθ), to φ. This can be considered
as a 'rigid' or global transformation that changes all points in the
field by the same amount. It is easy to see that the Lagrangian is
invariant under this transformation (φ' = φexp(iθ) and φ'^{*} = φ^{*}exp(-iθ)).
This implies there is a symmetry and a conserved quantity.
Now e^{iε} = 1 + iε ... Taylor series
Thus,
φ -> φ + iεφ -> φ + εf_{φ} where f_{φ} = iφ
φ^{*} -> φ^{*} - iεφ^{*} -> φ^{*} - εf_{φ*} where f_{φ*} = -iφ^{*}
Look for a conserved quantity using Noether's Theorem.
. .
π_{φ} = ∂L/∂φ = φ^{*}
. .
π_{φ*} = ∂L/∂φ^{*} = φ
Conserved quantity = ∫[π_{φ}f_{φ} + π_{φ*}f_{φ*}]d^{3}x
. .
= i∫[φ^{*}φ - φφ^{*}]d^{3}x
The quantity in [] is the charge density of the field, ρ.
Another way to write this is:
.
ρ = Im(φφ^{*})
∴ Q = ∫ρd^{3}x
We can also write a 4-vector for current, j^{μ}, as:
j^{μ} = Im(φ^{*}∂^{μ}φ)
The word 'charge' can have several different meanings. It can be the
electric charge but also the weak hypercharge. The take away from all of
this is that the rotational symmetry of a field is associated with
charge. THUS, THE U(1) SYMMETRY IS ASSOCIATED WITH CHARGE.
Notes:
- Every field has a field quanta. E-M field => photon.
- Every particle is a quanta of a field.
- CHARGED PARTICLES ARE ALWAYS DESCRIBED BY COMPLEX FIELDS.
- Don't confuse ρ with the e-m field. They are different things.
- Need to couple ρ with e-m field in a Lagrangian to see how they
interact.
Gauge Theory
------------
A symmetry is a transformation that preserves relevant physical properties
such as the equations of motion. It can be described as a transformation
that leaves the Lagrangian invariant. A GLOBAL symmetry is a symmetry
transformation that is executed by the same 'amount' at every point. A LOCAL
symmetry has a transformation that depends on space-time. In the case of
rotation, the angle of the rotation can change for different locations and times
A gauge symmetry is a local symmetry that relates states that are physically
the same. The states it connects are then only different in their mathematical
structure. The presence of a gauge symmetry can be viewed as a redundancy
in description. In the words there is more description and more mathematical
structure than is strictly necessary to describe the physical state. Gauge
transformations represent a stronger symmetry than the global symmetry
associated with a rigid transformation where θ is not a function of position.
We will see how the use of a gauge transformation can make an otherwise
invariant Lagrangian invariant.
Consider a complex field φ = φ_{R} + iφ_{I} and φ^{*} = φ_{R} - iφ_{I} and rotate by a phase
θ(x) to get φ' as follows:
φ' = φe^{iθ(x)} and φ'^{*} = φ^{*}e^{-iθ(x)}
This is a local U(1) symmetry.
Recall from the previous section on complex fields that:
L = ∂_{μ}φ'∂^{μ}φ'^{*} - m^{2}φ'φ'^{*}
After some math the Lagrangian density becomes (dropping the factor of 1/2 to
simplify matters):
(∂_{μ}φ - i∂_{μ}θφ)(∂_{μ}φ^{*} + i∂_{μ}θφ^{*}) - m^{2}φφ^{*}
The m^{2}φφ'^{*} term is invariant but the rest is not. In order to make the
Lagrangian invariant under the local U(1) symmetry, we need to add the
concept of the GAUGE COVARIANT DERIVATIVE, D_{μ}. For the electromagnetic
field D_{μ} is defined as:
D_{μ}φ = ∂_{μ}φ + iqA_{μ}φ and
D^{μ}φ^{*} = ∂^{μ}φ^{*} - iqA^{μ}φ^{*}
where A_{μ} is the GAUGE FIELD written as a 4-vector = (A_{0},A_{1},A_{2},A_{3}). In the
case of the electromagnetic field A_{0}, is the electrostatic potential and
A_{1..3} is the magnetic vector potential E = -∇A_{0} - ∂A_{1..3}/∂τ and B = ∇xA_{1..3}.
Note: A_{μ} itself is not itself gauge invariant. q is the charge (coupling
constant). The electric charge is just a measure of the strength of a
particle's interaction with the electromagnetic field. Saying that a
particle has a given charge is the same as saying it interacts with the
electromagnetic field with a certain strength.
In order for the Lagrangian density to be invariant it is necessary that
A transform as:
A_{μ}' = A_{μ} - (1/q)∂_{μ}θ
Proof:
D_{μ}φ' = ∂_{μ}φ' + iqA_{μ}'φ'
= (∂_{μ}φ + iφ∂θ/∂x)e^{iθ(x)} + iq(A_{μ}φ - (1/q)φ∂θ/∂x)e^{iθ(x)}
= (∂_{μ}φ + iqA_{μ}φ)e^{iθ(x)}
= e^{iθ(x)}D_{μ}φ
Similarly,
D^{μ}φ'^{*} = e^{-iθ(x)}D^{μ}φ^{*}
Therefore,
D_{μ}φ'D^{μ}φ'^{*} = D_{μ}φD^{μ}φ^{*} and the Lagrangian is gauge invariant.
We can rewrite the Lagrangian as:
L = D_{μ}φD^{μ}φ^{*} - m^{2}φφ^{*}
= (∂_{μ}φ + iqA_{μ}φ)(∂_{μ}φ^{*} - iqA^{μ}φ^{*})e^{iθ(x)}e^{-iθ(x)} - m^{2}φφ^{*}
= ∂_{μ}φ∂^{μ}φ^{*} + q^{2}A_{μ}A^{μ}φφ^{*} + iqA_{μ}[φ∂^{μ}φ^{*} - φ^{*}∂^{μ}φ] - m^{2}φφ^{*}
This Langrangian is invariant under the local U(1) symmetry.
The [] term represents a current and A_{μ}[] represents the
interaction between this current with the electromagnetic
field. Thus, we see that our theory which is invariant under
local gauge transformations is promoted to an interacting theory.
Invariance of A_{μ} by Itself
--------------------------
We have looked at how the introduction of the gauge field makes
φ invariant under a local transformation. However, the gauge
field itself must also have its own gauge invariant dynamic term
in the Lagrangian. Consider the construction, F_{μν}, defined as:
F_{μν} = ∂_{μ}A_{ν} - ∂_{ν}A_{μ}
Intuitively, this is a good choice since it contains the first
derivatives, ∂_{μ}, of the field that is characterisitic of all
Lagrangians.
If we plug in A_{μ}' = A_{μ} - (1/q)∂_{μ}θ into the above we get:
F_{μν} = ∂_{μ}A_{ν} + ∂_{μ}∂_{ν}θ - (∂_{ν}A_{μ} + ∂_{ν}∂_{μ}θ)
^{ }= ∂_{μ}A_{ν} - ∂_{ν}A_{μ}
Thus, under the gauge transformation, F_{μν}, remains unchanged. To make
this both gauge and Lorentz invariant and also match the quadratic
nature of the other terms in the Lagrangian, it is necessary to form
the product:
F_{μν}F^{μν}
We can now rewite our a Lagrangian for the gauge field as:
L = -F_{μν}F^{μν}
The is the Lagrangian that describes that electromagnetic field in
the absence of any charges (complex fields are charged fields so in
the absence of charge φ = 0).
The Higgs Mechanism
--------------------
In the previous section we have talked about the role of gauge fields
in constucting invariant Lagrangians. The quanta associated with
these fields are called GAUGE BOSONS. By analogy with the m^{2}φφ* term
associated with the complex field, we might expect that mass for the
gauge bosons might be associated with a m^{2}A_{μ}A^{μ} term. However, such
terms are not invariant and therefore break the local symmetry. The
invariance can be seen since m^{2}(A_{μ} + ∂_{μ}θ)(A^{μ} + ∂^{μ}θ) is clearly not equal
to m^{2}A_{μ}A^{μ}. The q^{2}A_{μ}A^{μ}φφ^{*} term from before represents an interaction term
and IS invariant because A_{μ} and A^{μ} have already been transformed during
the derivation. The consequence of this invariance is that, in general,
we can make the statement that gauge symmetry prohibits gauge bosons
from having mass.
The standard model is a gauge theory and particles do have mass. The
implication of this is that for mass to appear in the standard model,
the local symmetry has to be broken. This is referred to as SPONTANEOUS
SYMMETRY BREAKING.
Bosonic Mass
------------
Nearly all of the mass of protons and neutrons is from the kinetic energy
of their constituent quarks and gluons. This kinetic energy can only exist
if the quarks themselves have a rest mass. It is the Higgs field that is
responsible for giving mass to these constituent particles and also to the
electrons. The Higgs field is also responsible for giving mass to the REAL
W^{+}, W^{-} and Z^{0} gauge bosons that can be produced in collisions if sufficient
energy is available.
The potential energy, V(φ), of the Higgs field has the shape of a
'Mexican hat'. For ordinary fields the lowest energy is the point
corresponding to φ = 0. For the Higgs, however, the location of the
lowest energy is not at φ = 0 but at the point φ = f.
Any system, including a universe, will tumble into its lowest energy state,
like a ball bouncing down to the bottom of a valley. The universe, like a
ball, comes to rest somewhere on this circular trench, which corresponds
to a nonzero value of the field. That is, in its natural, lowest energy
state, the universe is permeated throughout by a nonzero Higgs field.
The Higgs field is a complex field represented by:
φ = φ_{R} + iφ_{I} and φ^{*} = φ_{R} - iφ_{I}
The equation representing the potential of the Higgs field is of the form:
V(φ) = -m^{2}φ^{2} + λφ^{4} where λ is a coupling constant.
Note: +m^{2}φ^{2} is a parabaloid centered at 0.
For the minima:
∴ dV(φ)/dφ = -2m^{2}φ + 4λφ^{3} = 0
so that,
φ = f = m/√2λ
f is referred to as the VACUUM EXPECTATION VALUE. The VEV for the
Higgs field is 246 GeV.
The Higgs potential can also be written as:
V(φ) = -m^{2}φφ^{*}+ λ(φφ^{*})^{2}
In this system, polar coordinates are much easier to work with than
cartesian coordinates. We can write:
φ = ρe^{iα} and φ^{*} = ρe^{-iα}
Where ρ represents the radial coordinate and α represents the
angular coordinate.
Abelian U(1) Gauge Theory
-------------------------
The particle at point A is symmetrical about rotation of the complex plane.
Thus, the Higgs field has U(1) symmetry. When this symmetry is spontaneously
broken, the particle 'descends' into the trough (point B) where the energy
is at a minimum. At this point the particle experiences a flat potential as
it moves along a circular path as shown. A movement in this direction does
not face any resistance since the energy in the adjacent state is the same.
A particle moving in this angular fashion therefore has zero mass. This is
the GOLDSTONE boson*. In contrast, oscillations back and forth along the
radial (red) direction are not massless. These are the HIGGS boson excitations.
Mathematically, we can get some insight into the Higgs mechanism as follows:
*Goldstone's Theorem: For each broken generator of the original symmetry
group one massless spin-zero particle will appear.
We can write the Lagrangian for a spin 0 field (the Higgs boson is spin 0
particle) as:
L = ∂_{μ}φ∂^{μ}φ^{*} - V(φ)
= (∂_{μ}φ_{R})^{2} + (∂_{μ}φ_{I})^{2} - V(φ)
In terms of ρ and α this is equivalent to:
L = (∂_{μ}ρ)^{2} + ρ^{2}(∂_{μ}α)^{2} - V(ρ)
Now ρ = f and is fixed and V(ρ) -> 0. Therefore, the Lagrangian can
be written as:
L = 0 + f^{2}(∂_{μ}α)^{2} - 0
= f^{2}(∂_{μ}α)^{2}
= (∂_{μ}fα)^{2}
This is characteristic of a field with quanta of zero mass. This is
the GOLDSTONE FIELD.
ρ represents the distance to the bottom of the trough, f, plus a
small displacement, H. Thus we can write:
ρ = f + H
The amount of energy required to excite the Higgs is quite large.
If we consider a low energy where the oscillations are small we
can write:
φ = fe^{iα}
Under a local U(1) transformation the field transforms as:
φ -> φexp(iθ(x))
However, this is not symmetric since ∂_{μ}φ∂^{μ}φ^{*} ≠ ∂_{μ}φ'∂^{μ}φ'^{*}. In order
to make it so we need to introduce the GAUGE FIELD, A_{μ} where A_{μ}
transforms as A_{μ} -> A_{μ} + ∂_{μ}θ and use the covariant derivative.
This results in:
D_{μ}φ = ∂_{μ}φ + iqA_{μ}φ
= i(∂_{μ}α + qA_{μ})fe^{iα}
Likewise,
D^{μ}φ^{*} = -i(∂_{μ}α + qA_{μ})fe^{-iα}
Therefore,
D_{μ}φD^{μ}φ^{*} = (qA_{μ} + ∂_{μ}α)^{2}f^{2}
^{ } = q^{2}f^{2}A_{μ}A^{μ} + 2qf^{2}A_{μ}∂_{μ}α + f^{2}(∂_{μ}α)^{2}
If we now make the UNITARY gauge transformation A_{μ} -> A_{μ} - (1/q)∂_{μ}α we get:
D_{μ}φD^{μ}φ^{*} = q^{2}f^{2}[A_{μ} - (1/q)∂_{μ}α][A^{μ} - (1/q)∂_{μ}α] + 2qf^{2}[A_{μ} - (1/q)∂_{μ}α]∂_{μ}α + f^{2}(∂_{μ}α)^{2}
^{ } = q^{2}f^{2}A_{μ}A^{μ} - 2qf^{2}A^{μ}∂_{μ}α + f^{2}(∂_{μ}α)^{2} + 2qf^{2}A_{μ}∂_{μ}α - 2f^{2}(∂_{μ}α)^{2} + f^{2}(∂_{μ}α)^{2}
^{ } = q^{2}f^{2}A_{μ}A^{μ}
^{ } ≡ m_{γ}^{2}A_{μ}A^{μ}
So,'mathemagically' the Goldstone boson disappears. This explains
why it is never seen in nature. In the process of disappearing, the
Goldstone boson has given rise to a mass m^{2}A^{2} equivalent to the mass
of a photon. It is instructive to count the degrees of freedom before
and after SSB has occurred. We started out with a massless photon
with 2 degrees of freedom and a complex scalar field with 2 degrees
of freedom for a total number of 4 degrees of freedom. After the
SSB we have one massive photon with 3 degrees of freedom and a real
scalar field with 1 degree of freedom, again for a total of 4 degrees
of freedom.
So this is a little confusing since we know the 'everyday' photon is
massless. We must conclude that spontaneous symmetry breaking does not
occur for the photon, or does it?. There is a special case where
spontaneous symmetry breaking does occur and that is within a
superconductor. In the theory of superconductivity the photon does
in fact have a mass.
So far we have assumed that the H in f + H is small and oscillations
of the field can be ignored. However, if the system is given enough
energy, the field will start oscillating in the radial direction along
ρ (the red curve). These oscillations represent the HIGGS BOSON.
As before, we can choose a direction of fluctuation so that the vacuum
Higgs field is:
φ = (f + H)e^{iα}
The Lagrangian now looks like:
L = D_{μ}{(f + H)e^{iα}}D^{μ}{(f + H)^{*}e^{-iα}} - m^{2}(f + H)^{2} - λ(f + H)^{4} - F_{μν}F^{μν}
= (∂_{μ} + iqA_{μ})(fe^{iα} + He^{iα})(∂^{μ} - iqA^{μ})(fe^{-iα} + He^{-iα}) - m^{2}(f + H)^{2}
- λ(f + H)^{4} - F_{μν}F^{μν}
After some math The Lagrangian becomes:
L = q^{2}f^{2}A_{μ}A^{μ} + 2qf^{2}A_{μ}∂_{μ}α + f^{2}(∂_{μ}α)^{2} + ∂_{μ}H∂^{μ}H - 4λf^{2}H^{2} - F_{μν}F^{μν}
+ cubic and quartic terms
This Lagrangian now describes a theory with a photon of mass m_{γ} = qf,
a Higgs boson, H, with m_{H} = 2√(λ)f and a massless Goldstone boson, α.
Again, we can make the UNITARY gauge transformation A_{μ} -> A_{μ} - (1/q)∂_{μ}α
to get:
L = q^{2}f^{2}A_{μ}A^{μ} + ∂_{μ}H∂^{μ}H - 4λf^{2}H^{2} - F_{μν}F^{μν} + cubic and quartic terms
Now there are no massless particles in this theory, and the α field has
completely disappeared from the Lagrangian.
Thus, when the Higgs field is excited, the Higgs boson also get its
mass as a consequence of spontaneous symmetry breaking.
Note: It is inportant to distinguish between the Higgs field and the
Higgs boson. The Higgs field is the entity that gives all other particles
a mass. The Higgs boson is a particle (quanta of the field) that gets
its mass like all other particles. Once the Higgs particle has been
created, it will eventually decay. Though the Higgs particle interacts
with all massive particles it prefers to interact with the heaviest
elementary particles. The Higgs particle is considered to be a carrier
of a force. It is a boson, like the other force-transferring particles:
photons, gluons and electroweak bosons. It has 0 spin, carries no
charge and has a mass of about 126 GeV.
Non-Abelian SU(2)⊗U(1) Gauge Theory (Yang-Mills)
------------------------------------------------
In the interest of simplicity, the above arguments have only considered
the Abelian U(1) gauge symmetry breaking and are by no means rigorous.
The discussion has been focussed around the photon and the mechanism for
gaining mass. The sole purpose has been only to provide a basic idea of
the fundamental concepts and in this respect the Abelian U(1) model is
sufficient.
For the SU(2) symmetry that occurs in the electroweak theory of the
Standard Model, the elements in the Lagrangian are a bit different.
The gauge symmetric theory for SU(2) was developed by YANG and
MILLS. Under a local SU(2) transformation the field transforms as:
φ -> φexp(iθ_{i}(x)σ_{i}/2)
where σ_{i} are the Pauli matrices - the generators of SU(2). Note:
These are NOT the spin angular momentum generators. They are, instead,
asscociated with the weak isospin. The physical interpretation is
completely different, but the algebra is identical to that of angular
momentum. θ_{i} are the 3 group parameters.
The covariant derivative for a SU(2)⊗U(1) gauge invariant Lagrangian
becomes:
D_{μ} = ∂_{μ} + igW_{μ}^{i}σ_{i}/2 + ig'Y_{W}B_{μ}
Where g and g' are coupling constants.
The W_{μ}^{i} fields are associated with the weak isospin, the generator of
SU(2). Since there are 3 generators, there are 3 gauge fields (with
massless bosons) designated by W_{μ}^{0}, W_{μ}^{1} and W_{μ}^{2}. Likewise the B_{μ} field
is associated with the weak hypercharge, Y_{W}, the generator of U(1). Note
that this U(1) is NOT the same as the U(1) in the Abelian case discussed
above for the photon. In that case, U(1) is associated with the
conventional charge, q. By analogy with photons, particles with
weak hypercharge interact by the exchange of B bosons - the massless
quanta of the B gauge field.
Due to the non-commutativity of the generators of this symmetry
([σ_{1},σ_{2}] = iσ_{3} etc.), the gauge tensor for the W fields is:
G_{μν}^{i} = ∂_{μ}W_{ν}^{i} - ∂_{ν}W_{μ}^{i} + f^{ijk}W_{μ}^{j}W_{ν}^{k}
where f^{ijk} is the structure constant of the particular gauge group.
The basic form of the Langrangian that encapsulates SU(2)⊗U(1) looks
like (omitting constants):
L = D_{μ}φD^{μ}φ^{†} - V(φ) - F_{μν}F^{μν} - G_{μν}^{i}G^{μν}_{i}
where F_{μν} has been redefined as:
F_{μν} = ∂_{μ}B_{ν} - ∂_{ν}B_{μ}
In the Abelian case the Higgs field was considered to be a complex
scalar field with 2 components. In the non-Abelian case the Higgs
field is a complex scalar field of the group SU(2) with 4 components
- 2 neutral (0) and 2 charged (Q) each with 1 degree of freedom for
a total of 4. Neglecting normalization factors, the field is a
weak isospin doublet written as:
- - - - - -
φ = | φ^{+} | = | φ_{1} + iφ_{2} | ≡ | H^{+} + iH^{-} | with I_{3} = 1/2 and Y_{W} = 1/2
| φ^{0} | | φ_{3} + iφ_{4} | | H^{ } + iH ^{0}| with I_{3} = -1/2 and Y_{W} = 1/2
- - - - - -
and its Hermitian conjugate,
- -
φ^{†} = | φ^{-} φ^{0*} |
- -
Under U(1) rotations, these doublets are multiplied by a phase exp(iφY_{W})
Gauge symmetry is spontaneously broken down to the U(1) symmetry of
electromagnetism. This can be written as:
SU(2)_{L}⊗U(1)_{Y} -> U(1)_{EM}
Where L denotes left-handed chirality, Y_{W} denotes the weak hypercharge
and EM denotes the electromagnetic field.
As in the Abelian case, for low energies (H very small) we can write:
D_{μ}φD^{μ}φ^{†} = (∂_{μ}φ + igW_{μ}^{i}σ_{i}φ/2 + ig'Y_{W}B_{μ}φ)(∂_{μ}φ^{†} - igW_{μ}^{i}σ_{i}φ^{†}/2 - ig'Y_{W}B_{μ}φ^{†})
After some math and, for the purpose of illustration, neglecting
all constants, we get:
D_{μ}φD^{μ}φ^{†} = f^{2}[g^{2}{(W_{μ}^{1})^{2} + (W_{μ}^{2})^{2}} + (gW_{μ}^{0} - g'B_{μ})^{2}]
The g^{2} term can be written as W_{μ}W_{μ}^{†} where W_{μ} and W_{μ}^{†} are LADDER
operators defined as:
W_{μ} = W_{μ}^{1} + iW_{μ}^{2} = W^{-}
W_{μ}^{†} = W_{μ}^{1} - iW_{μ}^{2} = W^{+}
Ladder operators were introduced in the context of angular momentum
and rotation so it is also appropriate to use them in the context of
rotations associated with SU(2).
Thus, we can write:
f^{2}g^{2}{(W_{μ}^{1})^{2} + (W_{μ}^{2})^{2}} = f^{2}g^{2}W_{μ}W_{μ}^{†} and identify the mass m_{W} = gf
The 2 remaining neutral Z and γ bosons are defined in terms of the
WEINBERG ANGLE, θ_{W}, as follows:
Z_{μ}^{0} = W_{μ}^{0}cosθ_{W} - B_{μ}sinθ_{W}
γ_{μ} = W_{μ}^{0}sinθ_{W} + B_{μ}cosθ_{W}
θ_{W} is found experimentally and is the angle by which SSB rotates the
original W_{μ}^{0} and B_{μ} vector boson plane. It is defined as:
cosθ_{W} = g/√(g^{2} + g'^{2})
sinθ_{W} = g'/√(g^{2} + g'^{2})
It also gives the relationship between the masses of the W and Z
bosons as follows:
m_{Z} = m_{W}/cosθ_{W}
Therefore,
Z_{μ}^{0} = gW_{μ}^{0}/√(g^{2} + g'^{2}) - g'B_{μ}/√(g^{2} + g'^{2})
^{ } = (1/√(g^{2} + g'^{2}))(gW_{μ}^{0} - g'B_{μ})
with mass, m_{Z} = m_{W}/g/√(g^{2} + g'^{2}) = gf/g/√(g^{2} + g'^{2}) = f√(g^{2} + g'^{2})
and
γ_{μ} = g'W_{μ}^{0}/√(g^{2} + g'^{2}) + gB_{μ}/√(g^{2} + g'^{2})
^{ } = (1/√(g^{2} + g'^{2}))(g'W_{μ}^{0} + gB_{μ})
with mass m_{γ} = 0
It is instructive to look at the degrees of freedom before and after SSB
has occurred. At the outset there are 4 degrees of freedom associated
with the Higgs field, 2 associated with the massless B field and 6
associated with the 3 massless W fields for a total of 12. After SSB
there is a real scalar field H with 1 degree of freedom, 3 massive weak
bosons, W^{±} and Z^{0}, with 9, and 1 massless photon with 2, again yielding
a total of 12. The scalar field degrees of freedom have been 'eaten'
to give the W^{±} and Z^{0} bosons their longitudinal components. Note:
massless particles cannot have a degree of freedom in their longitudinal
direction since they are constrained to travel at the speed of light
along their direction of motion.
Schematically, the before and after SSB scenario is illustrated below:
- - - -
| H^{+} + iH^{-} | -> | 0 |
| H^{ } + iH^{0} | | f + H |
- - - -
Under infinitesinal rotations the vacuum, φ_{0}, should be symmetric.
Therefore,
φ_{0} -> exp(iθZ)φ_{0} where Z is the rotation.
This means that,
φ_{0} -> (1 + iθZ)φ_{0}
So Zφ_{0} must equal 0.
SU(2)_{L}:
- - - - - -
σ_{3}φ_{0} = | 1 0 || 0 | = -| 0 | ≠ 0 -> broken
_{ } | 0 -1 || f + H | | f + H |
- - - - - -
U(1)_{Y}:
- - - - - -
Y_{W}φ_{0} = | 1 0 || 0 | = | 0 | ≠ 0 -> broken
_{ } | 0 1 || f + H | | f + H |
- - - - - -
U(1)_{EM}:
- - - -
Qφ_{0} = (1/2)(σ_{3} + Y_{W}) = | 1 0 || 0 | = 0 -> unbroken
_{ } | 0 0 || f + H |
- - - -
After SSB, the H^{+}, H^{-} and H^{0} components all become Goldstone bosons.
The 4th component, H, gets a VEV, f. The 3 Goldstone bosons get 'eaten'
by combinations of the massless W_{μ}^{0}, W_{μ}^{1}, W_{μ}^{2} and B_{μ} field bosons in the
following way:
The negative combination of W_{μ}^{1} and W_{μ}^{2} 'eats' H^{+} giving the W^{+} boson
a longitudinal 3rd polarization (i.e. mass).
Likewise, the positive combination of W_{μ}^{1} and W_{μ}^{2} 'eats' H^{-} to give the
W^{-} boson a mass.
And the negative combination of W_{μ}^{0} and B_{μ} 'eats' H^{0} to give the
Z^{0} boson a mass.
Finally, the leftover combination of W_{μ}^{0} and B_{μ} forms the photon. But
since there are no more Goldstone bosons to 'eat' the photon remains
massless.
We could repeat this process with (f + H) to see the involvement of
the the Higgs boson. However, the math is more cumbersome and
therefore is not attempted here. The end result is the Higgs boson
component forms a condensate which, as we will see, can interact
with fermions to also give them mass.
SU(3) ⊗ SU(2) ⊗ U(1) Symmetry Breaking
---------------------------------------
Under a local SU(3) transformation the field transforms as:
φ -> φexp(iθ_{i}(x)λ_{i}/2)
where λ_{i} are the Gell-Mann matrices - the generators of SU(3). Since
there are 8 generators there are 8 gauge fields designated by G_{μ}^{i} which
correspond to the gluons.
The covariant derivative for a SU(3)⊗SU(2)⊗U(1) gauge invariant Lagrangian
is:
D_{μ} = ∂_{μ} + ig''G_{μ}^{i}λ_{i}/2 + igW_{μ}^{i}σ_{i}/2 + ig'Y_{W}B_{μ}
Where g, g' and g'' are coupling constants.
Due to the non-commutativity of the generators of this symmetry, the
gauge tensor is:
G_{μν} = ∂_{μ}G_{ν}^{i} - ∂_{ν}G_{μ}^{i} + f^{ijk}G_{μ}^{j}G_{ν}^{k}
where f^{ijk} is the structure constant of the particular gauge group.
Fermionic Mass
--------------
- -
φ = | 0 | e^{iα}
| f + H |
- -
- -
φ^{†} = | 0 f + H | e^{-iα}
- -
The fermions also get their mass from the spontaneous symmetry breaking
of the Higgs Field. Fermions include quarks, neutrinos, electrons, muons
and tau's.
It has been experimentally discovered that in the decay of a neutron to
a proton, the electron emitted is always left handed. This turns out to
be true for any process involving the weak interaction. For example,
an interaction involving a W boson and an electron to produce a neutrino,
will only take place if the electron is left handed. In addition, the
neutrino produced will always be left handed.
For any fermion we can write the Dirac equation in terms of right
and left handed particles as follows (h = c = 1):
i∂ψ_{R}/∂t + iα_{i}∂ψ_{R}/∂x = mψ_{L}
i∂ψ_{L}/∂t + iα_{i}∂ψ_{L}/∂x = mψ_{R}
Lets consider the electron. The mass terms combine two particles, e_{R} and
e_{L} into a single particle - the 'physical electron' which is propagating
through space. e_{L} and e_{R} swap back and forth. At a particular point
in time the particle may be e_{L}, but if you observe it a moment later,
the very same particle might manifest itself as e_{R}. Weak interactions
with the 'physical electron' occur through its left-handed component
and not with the right-handed component.
At this point we need to consider the weak hypercharge associated with
the weak interaction. The weak hypercharge, Y_{W}, relates the electric
charge, Q, and I_{3}. It satisfies the equality:
Q = I_{3} + Y_{W}/2
Where Q is the charge and I_{3} is the third component of the isospin.
∴ Y_{W} = 2(Q - I_{3})
Consider the case for electrons. Y_{W} for a left-handed electron, e_{L},
is -1 and for a right-handed electron, e_{R}, is -2. It is clear that,
as it stands, plugging these values into the 2 Dirac equations
results in a charge violation that 'breaks' the coupling between
the two versions of the electron.
Now, the Higgs field, H, has a weak hypercharge equal to 1.
If we rewrite the Dirac Equations as:
i∂ψ_{R}/∂t + iα_{i}∂ψ_{R}/∂x = mH^{†}ψ_{L}
i∂ψ_{L}/∂t + iα_{i}∂ψ_{L}/∂x = mHψ_{R}
Under SSB, the weak hypercharge is now conserved. In essence, the
Higgs field become a condensate that acts as a source and sink of
the weak hypercharge. The addition/subtraction of H (f) can be
looked at in the following way:
/H (provides weak hypercharge)
/
rh------g_{y}
\
\ lh
/H^{†}(removes weak hypercharge)
/
lh------g_{y}
\
\ rh
Now, H = f (the VEV) and we can write the 2 Dirac equations as (f^{†} = f):
i∂ψ_{R}/∂t + iα_{i}∂ψ_{R}/∂x = g_{y}fψ_{L}
i∂ψ_{L}/∂t - iα_{i}∂ψ_{L}/∂x = g_{y}fψ_{R}
where g_{y} is the YUKAWA coupling constant and represents the interaction
between the Higgs field and a Dirac field.
It is apparent from this the mass term in the original equations
is equivalent to g_{y}f. Therefore, we can write:
m = g_{y}f
There is a g_{y} for every type of fermion. Example, for an electron -
W boson interaction:
g_{y} ~ 90GeV/0.5Mev ~ 10^{-5}
The above calculations have assumed the low energy case where we
ignored the oscillations of the Higgs field and assumed that f is
a constant. To include the Higgs boson we would need to rewrite
the right hand side of the equations as follows:
i∂ψ_{R}/∂t + iα_{i}∂ψ_{R}/∂x = g_{y}(f + H)ψ_{L}
i∂ψ_{L}/∂t + iα_{i}∂ψ_{L}/∂x = g_{y}(f + H)ψ_{R}
Alternative Derivation
----------------------
We can also derive similar results from the Lagrangian of the Dirac
equation. The Dirac field has 4 components: Positive and negative
energies and left and right handedness (spin).
The Dirac equation with h = c = 1 is:
(iγ^{μ}∂_{μ} - m)ψ = 0
Solutions to this equation are the DIRAC SPINORS.
_ _
To get the Lagrangian multiply by ψ where ψ = γ^{0}ψ^{†} ... the DIRAC
ADJOINT SPINOR. Thus,
_
L = ψ(iγ^{μ}∂_{μ} - m)ψ
_
ψ is interpreted as a creation operator and ψ is interpreted as an
annihilation operator.
Note: We can prove that this is the Lagrangian by applying the
Euler-Lagrange equations. The result yields the original Dirac
equation. Also, this is not the Lagrangian for QED. The QED
Lagrange contains D_{μ} and the electromagnetic
tensor.
- - _{ } - - - -
ψ = | ψ_{1} | ψ^{†} = | ψ_{1}^{†} ψ_{2}^{†} ψ_{3}^{†} ψ_{4}^{†} | γ^{0}ψ^{†} = | ψ_{1}^{†} ψ_{2}^{†} -ψ_{3}^{†} -ψ_{4}^{†} |
| ψ_{2} | _{ } - - - -
| ψ_{3} |
| ψ_{4} |
- -
The purpose of γ^{0} is to make the Lagrangian Lorentz invariant.
Rewrite the DE as:
_ ^{ } _
iψγ^{μ}∂_{μ}ψ - mψψ = 0
The first term in the Lagrangian is the kinetic term. The second term
flips the chirality back and forth. We can write:
The Dirac spinor, ψ, can be written as a linear combinations of
left-handed and right-handed states. Therefore,
ψ = ψ_{L} + ψ_{R} and ψ^{†} = ψ_{L}^{†} + ψ_{R}^{†}
_
Thus, using the relationship ψ = ψ^{†}γ^{0}, we get:
_
ψψ = ψ^{†}γ^{0}ψ
- - - - - -
= | ψ^{†}_{L} ψ^{†}_{R} || 0 1 || ψ_{L} |
- - | 1 0 || ψ_{R} |
- - - -
= ψ^{†}_{L}ψ_{R} + ψ^{†}_{R}ψ_{L}
The first term of the Lagrangian can be written as:
L_{KE} = iψ^{†}_{L}γ^{μ}∂_{μ}ψ_{L} + iψ^{†}_{R}γ^{μ}∂_{μ}ψ_{R}
In order to account for the interaction of the fermions with the
symmetry group SU(2)⊗U(1) and the gauge fields we have to replace
∂_{μ} with the gauge covariant derivative, D_{μ}. Thus,
L_{KE} = iψ^{†}_{L}γ^{μ}D_{μ}ψ_{L} + iψ^{†}_{R}γ^{μ}D_{μ}ψ_{R}
The second term of the Lagrangian can be written as:
_
mψψ = mψ^{†}_{L}ψ_{R} + mψ^{†}_{R}ψ_{L}
However, this as it stands would not be a legal term in the Lagrangian
because it violates weak hypercharge conservation. As before, we can
restore the coupling if we introduce the Higgs field, H, as follows:
_
g_{y}ψHψ = g_{y}(ψ^{†}_{L}Hψ_{R} + ψ^{†}_{R}H^{†}ψ_{L})
_
where g_{y}ψHψ is the YUKAWA INTERACTION
Consider the case H = f. We get:
_ _{ }
g_{y}ψfψ = g_{y}f(ψ^{†}_{L}ψ_{R} + ψ^{†}_{R}ψ_{L})
Now consider f + H. We get:
_ _{ }
g_{y}ψ(f + H)ψ = g_{y}(ψ^{†}_{L}(f + H)ψ_{R} + ψ^{†}_{R}(f + H)ψ_{L})
So the RHS becomes:
g_{y}f(ψ^{†}_{L}ψ_{R} + ψ^{†}_{R}ψ_{L}) + g_{y}H(ψ^{†}_{L}ψ_{R} + ψ^{†}_{R}ψ_{L}) + the hermitian conjugate
The first term is the fermion mass. The second term describes a
process whereby a fermion is absorbed and re-emitted alongside a
Higgs boson.
The Complete Electroweak Lagrangian
-----------------------------------
Before SSB:
L_{EW} = L_{Higgs} + L_{Gauge} + L_{Dirac} + L_{Yukawa}
= D_{μ}φD^{μ}φ^{†} - V(φ) - F_{μν}F^{μν} - G_{μν}^{i}G^{μν}_{i} + iψ_{L}γ^{μ}D_{μ}ψ_{L} + iψ_{R}γ^{μ}D_{μ}ψ_{R}
+ g_{y}(ψ_{L}φψ_{R} + ψ_{R}φ^{†}ψ_{L})
Majorana Mass
-------------
The kind of fermion mass that we discussed above is called a Dirac mass.
This is a type of mass that connects two different particles, e_{R} and e_{L}.
It is also possible to have a mass that connects two of the same kind of
particle, this is called a MAJORANA mass.
This type of mass is forbidden for particles that have any type of charge.
There is, however, one type of matter particle in the Standard Model which
does not carry any charge - the neutrino. Majorana masses mix neutrinos
with anti-neutrinos so that the “physical neutrino” is its own antiparticle.